Question

If $I_1={\int }_0^{\frac{\pi }{2}}f(sin2x)sinxdx$ and $I_2={\int }_0^{\pi /4}f(cos2x)cosxdxthen{I}_1/I_2$ =

• A. $1$
• B. $\sqrt{2}$
• C. $\frac{1}{\sqrt{2}}$
• D. $2$

Answer 1

Answer: (B)

Given $I_1=\underset{0}{\overset{\pi /2}{\int }}f(sin2x)sinxdx$
$\Rightarrow I_1=\underset{0}{\overset{\pi /2}{\int }}f(sin2x)cosxdx$
$(\because \underset{0}{\overset{a}{\int }}f(x)dx=\underset{0}{\overset{a}{\int }}f(a-x)dx]$
$\Rightarrow 2I_1=\underset{0}{\overset{\pi /2}{\int }}f(sin2x)(sinx)+cosx)dx$
$=\sqrt{2}\underset{0}{\overset{\pi /2}{\int }}f(sin2x)cos(x-\frac{\pi }{4})dx$
Put $x-\frac{\pi }{4}=t$
$\Rightarrow dx=dt$
$\therefore 2I_1=\sqrt{2}\underset{-\pi /4}{\overset{\pi /4}{\int }}f(sin(\frac{\pi }{2}+2t))costdt$
$\therefore 2I_1=2\sqrt{2}\underset{0}{\overset{\pi /4}{\int }}f(cos2t)costdt$
$\Rightarrow I_1=\sqrt{2}\underset{0}{\overset{\pi /4}{\int }}f(cos2x)cosxdx$
$\Rightarrow I_1=\sqrt{2}{I}_2$